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If P != NP, are there then more Polynomial problems than SuperPolynomial problems, or vice versa?

templatetypedef
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Albert Hendriks
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    This question appears to be off-topic because it is about theory of computation, and not about a specific programming problem. Try http://cs.stackexchange.com/, perhaps. – amalloy Dec 13 '14 at 12:03
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    @Gumbo They wouldn't appreciate this question at CSTheory, which is for researchers in the field and not for the more casual questioner - Computer Science is the place. – amalloy Dec 13 '14 at 12:05
  • This question appears to be off-topic because it is about CS theory, but not research-level CS theory, and therefore should probably be migrated to cs.stackexchange.com. – templatetypedef Dec 16 '14 at 02:15
  • @simonzack I think that it's tricky to look for answers to this question. It's hard to find anything about the cardinality of P, NP, etc. because it's so rarely discussed (and it's so rarely discussed because many people consider it to be pretty uninteresting.) – templatetypedef Dec 16 '14 at 02:16

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From a formal languages perspective, there are only countably many problems in P and uncountably many problems not in P. Every problem in P can be solved by a deterministic, polynomial-time Turing machine, and since the number of TMs is countably infinite, the number of languages in P is countably infinite. On the other hand, the number of total languages is equal to the number of possible subsets of strings, so there are uncountably many languages not in P. This result is, interestingly enough, independent of whether P = NP.

If you restrict "problems" to "decidable problems" (that is, problems that are solvable by computers with unbounded time and storage space), then we know that there are only countably many total decidable problems. Countably infinitely many of them are in P and, regardless of whether P = NP, there are countably infinitely many of them not in P.

Hope this helps!

templatetypedef
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  • Your answer seems correct given my question, but I meant something else. Sorry about that. I meant P problems versus superPolynomial problems in NP. – Albert Hendriks Dec 16 '14 at 09:06
  • @AlbertHendriks Oh, it's the same number again. You can construct countably infinitely many NP-complete problems, none of which will be in P if P != NP. – templatetypedef Dec 16 '14 at 17:36